Let $\lVert \cdot \rVert$ denote a norm on $\mathbb{C}^m$. Define the dual norm $\lVert \cdot \rVert '$ by $\lVert x\rVert' := \sup_{\lVert y\rVert = 1} |y^* x|,$ where $y^*$ is the conjugate transpose of $y$. Let $x,y\in \mathbb{C}^m$ where $\lVert x\rVert =\lVert y\rVert = 1$. Show there is a rank-one matrix B so that $Bx = y$ and $\lVert B \rVert = 1$, where $\lVert B\rVert$ is the matrix norm of $B$ induced by the vector norm.
It could be useful to use the following fact, which I'd also like to know a proof for: given
$x\in\mathbb{C}^m,\exists z\neq 0\in\mathbb{C}^m$ so that $|z^*x| = \lVert z\rVert' \lVert x\rVert$.
Using the fact, we can let $z_0$ be a vector satisfying the fact and then I think we can let $B = yz^*$ where $z = e^{-\textrm{i}\theta} z_0/\lVert z_0\rVert'$ and $\theta$ is the argument of $z_0^*x$. Then $$Bx = y(z^*x) = z_0^*x e^{-i\theta}/\lVert z_0\rVert' = y.$$
But I'm not sure how to show that $\lVert B\rVert := \sup_{\lVert u\rVert = 1} \lVert Bu\rVert = 1$ since it may not be true that $Bu = y$ for all vectors $u$ with norm $1$.
Best Answer
Let $x, y \in \mathbb{C}^n$ and assume that $x \not = 0$. We shall show that there exists a matrix $B$ of rank at most $1$ such that $y = Bx$. Let $B$ be given by $$B = \frac{1}{\|x\|_2^2} yx^*.$$ Then $B$ has rank $1$ if and only if $y \not = 0$. Moreover, we have $$Bx = \frac{1}{\|x\|^2} y x^* x = y,$$ because, by definition, $$\|x\|_2^2 = x^*x.$$ We shall now determine the 2-norm of matrix $B$. Let $z \in \mathbb{C}^n$ be given. Then $$Bz = \frac{1}{\|x\|_2^2} y x^*z.$$ We now apply Cauchy-Schwartz's inequality and the fact that the matrix norm is submultiplicative. We deduce that $$\|Bz\|_2 \leq \frac{1}{\|x\|_2} \|y\|_2 \|x^*\|_2 \|z\|_2.$$ We conclude that $$\|B\|_2 \leq \frac{1}{\|x\|_2} \|y\| \|x^*\| = \frac{\|y\|_2}{\|x\|_2}.$$ It is important to appreciate the subtle fact that the 2-norm of the vector $x$ is equal to the 2-norm of the matrix $x^*$. This follows from Cauchy-Schwartz's inequality.
We now claim that $$\|B\|_2 = \frac{\|y\|_2}{\|x\|_2}.$$ To this end we examine $Bx$. We have $Bx = y$, so $$\|Bx\|_2 = \|y\|_2 = \frac{\|y\|_2}{\|x\|_2} \|x\|_2.$$ We can now conclude that $\|B\|_2 = \|y\|_2/\|x\|_2.$
In the original problem, $\|x\|_2 = \|y\|_2 = 1$. It follows that $B = yx^*$ has rank $1$ and $\|B\|_2 = 1$.